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3 years ago

What are considered measures of variability

1 answer:

7 0

Statisticians use summary measures to describe the amount of variability or spread in a set of data. The most common measures of variability are the range, theinterquartile range (IQR), variance, and standard deviation. This is from google btw

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The image of C reflected across the line x = 3 would have the coordinates ________________.

kipiarov [429]

Using translation concepts, the image of C reflected across the line x = 3 would have the coordinates (4,-2).

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.

The rule for a reflection across the line x = 3 is given by:

(x,y) -> (3 + |x - 3| , y).

The coordinates of C are C(2,-2), hence:

(2,-2) -> (3 + |2 - 3| , -2) = (4,-2).

The image of C reflected across the line x = 3 would have the coordinates (4,-2).

More can be learned about translation concepts at brainly.com/question/4521517

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3 0

2 years ago

Plz with steps .. it's very hard can anyone plz

liubo4ka [24]

Answer:

Step-by-step explanation:

$\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a} \qquad f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a} \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} }{\dfrac{1}{2\sqrt{x}} }\\\\$

$\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}} =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\$

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3 years ago

What is the area of a circle with a radius of 6 inches? (Leave your answer in terms of Pi. This means: do not multiple by 3.14,

shusha [124]

12 pi inches square

4 0

3 years ago

Given: RS = 3x - 16 ST = 4x - 8 RT = 60 Solve for RS.

Aleks04 [339]

The value of RS in the line segment is 20

To solve for RS :

From the line segment attached :

The information given is :

RS = 3x - 16

ST = 4x - 8

RT = 60

RT = RS + ST

60 = 3x - 16 + 4x - 8

60 = 7x - 24

60 + 24 = 7x

84 = 7x

Divide both sides by 7

84/7 = x

x = 12

RS = 3x - 16

RS = 3(12) - 16

RS = 36 - 16

RS = 20

HENCE, RS = 20

Learn more : brainly.com/question/24622406

6 0

3 years ago

John has purchased collision insurance with a $100-deductible clause, and automobile medical payments insurance in the amount of

stepladder [879]

He would have to pay them 1000 cause after he gets his car fix he will have 2000 left

4 0

4 years ago